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Looking for some help in proving why the following infinity goes to $\infty$ as n approaches $\infty$

$$\frac{(1-\frac{1}{n})^n}{1-(1-\frac{1}{n})}$$

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    Well, the numerator $\to 1/e$ and the denominator $\to 0$, ... – 2017-02-17
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    so this limit dne? – 2017-02-17
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    Depends on who you talk to. Some people say a limit DNE if the limit is infinity, others say it does exist and it's infinity. – 2017-02-17

2 Answers 2

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HINT:

So, we have $\lim_{n\to\infty}n\cdot\left[\lim_{n\to\infty}\left(1-\dfrac1n\right)^{-n}\right]^{-1}$

Do you know $\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$

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    yes I have learned that it does equal e – 2017-02-17
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    how did you change the limit around like that? – 2017-02-17
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    @jh123, $$\lim_{x\to a} f(x)g(x)=\lim_{x\to a} f(x) \lim_{x\to a} g(x) $$ provided at least one of them limit is non-zero and finite – 2017-02-17
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    so then $(e)(n)$ goes to infinity as n goes to infinity? – 2017-02-17
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    @jh123, Its $$\dfrac ne$$ – 2017-02-17
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    could you edit that into your answer? – 2017-02-17
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    @jh123, I shouldn't as it's hint – 2017-02-17
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$$\dfrac{(1-\dfrac{1}{n})^n}{1-(1-\dfrac{1}{n})}=\\ \dfrac{\binom{n}{0}1^{n}{(-\dfrac{1}{n})}^{0}+\binom{n}{1}1^{n-1}{(-\dfrac{1}{n})}^{1}+\binom{n}{2}1^{n}{(-\dfrac{1}{n-2})}^{2}+...+\binom{n}{n}1^{0}{(-\dfrac{1}{n})}^{n} }{1-(1-\dfrac{1}{n})}=\\ \dfrac{1+n{(-\dfrac{1}{n})}+\dfrac{n(n-1)}{2}{(-\dfrac{1}{n})}^{2}+\dfrac{n(n-1)(n-2)}{6}{(-\dfrac{1}{n})}^{3}+...+1{(-\dfrac{1}{n})}^{n} }{1-(1-\dfrac{1}{n})}=\\ \dfrac{1-1+\dfrac{(n-1)}{2}{(\dfrac{1}{n})}+\dfrac{(n-1)(n-2)}{6}{(-\dfrac{1}{n^2})}+...+1{(-\dfrac{1}{n})}^{n} }{\dfrac{1}{n}}=\\$$ so apply limit $$\lim_{n \to \infty }\dfrac{1-1+\dfrac{(n-1)}{2}{(\dfrac{1}{n})}+\dfrac{(n-1)(n-2)}{6}{(-\dfrac{1}{n^2})}+...+1{(-\dfrac{1}{n})}^{n} }{\dfrac{1}{n}}=\\ \lim_{n \to \infty }\dfrac{\dfrac{(n-1)}{2n}-\dfrac{(n-1)(n-2)}{6n^2}+...+1{(-\dfrac{1}{n})}^{n} }{\dfrac{1}{n}}=\\ \lim_{n \to \infty }n({\dfrac12-\dfrac16+\dfrac{1}{24}-\dfrac{1}{120}) }=\to \infty\\ $$